Lời giải
Đề bài:
Chứng minh rằng:$1+\frac{1}{2}C^{1}_{n}+\frac{1}{3}C^{2}_{n}+…+\frac{1}{n+1}C^{n}_{n}
Lời giải
Xét: $f(x)=(1+x)^{n}=\sum\limits_{k=0}^n C^{k}_{n}1^{n-k}x^{k}$
$=\sum\limits_{k=0}^n C^{k}_{n}x^{k}=C^{0}_{n}+C^{1}_{n}x+C^{2}_{n}x^{2}+…+C^{n}_{n}x^{n}$
$\Rightarrow \int\limits_{0}^{1}f(x)dx=\int\limits_{0}^{1}C^{0}_{n}dx+\int\limits_{0}^{1}C^{1}_{n}xdx+…+\int\limits_{0}^{1}C^{n}_{n}x^{n}dx$
$\Rightarrow \frac{2^{n+1}-1}{n+1}=C^{0}_{n}+\frac{1}{2}C^{1}_{n}+\frac{1}{3}C^{2}_{n}+…+\frac{1}{n+1}C^{n}_{n}$
$\Rightarrow 1+\frac{1}{2}C^{1}_{n}+\frac{1}{3}C^{2}_{n}+…+\frac{1}{n+1}C^{n}_{n}=\frac{2^{n+1}-1}{n+1}$\Rightarrow $ (ĐPCM)
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